# Inverse fourier transform using tables

• Sep 29th 2011, 04:58 AM
math2011
Inverse fourier transform using tables
By using tables find the inverse Fourier transform of $\displaystyle \displaystyle e^{-\frac{k^2}{4} + \frac{2}{3} k}$.

The definition of inverse Fourier transform of $\displaystyle F(k)$ in my lecture notes is $\displaystyle f(x) = \mathcal{F}^{-1} \{ F(k) \} = \frac{1}{\sqrt{2\pi}} \int^\infty_{-\infty} e^{-ikx} F(k) dk$.

I guess this would involve convolution as $\displaystyle \mathcal{F}^{-1} \{ e^{-\frac{k^2}{4}} \} = \sqrt{2} e^{-x^2}$. What is $\displaystyle \mathcal{F}^{-1} \{ e^{\frac{2}{3} k} \}$? How do I get this from the Fourier transforms table?

The given solution is $\displaystyle \sqrt{2} e^{\frac{4}{9}} e^{-i \frac{4}{3} x - x^2}$.
• Sep 29th 2011, 06:09 AM
CaptainBlack
Re: Inverse fourier transform using tables
Quote:

Originally Posted by math2011
By using tables find the inverse Fourier transform of $\displaystyle \displaystyle e^{-\frac{k^2}{4} + \frac{2}{3} k}$.

The definition of inverse Fourier transform of $\displaystyle F(k)$ in my lecture notes is $\displaystyle f(x) = \mathcal{F}^{-1} \{ F(k) \} = \frac{1}{\sqrt{2\pi}} \int^\infty_{-\infty} e^{-ikx} F(k) dk$.

I guess this would involve convolution as $\displaystyle \mathcal{F}^{-1} \{ e^{-\frac{k^2}{4}} \} = \sqrt{2} e^{-x^2}$. What is $\displaystyle \mathcal{F}^{-1} \{ e^{\frac{2}{3} k} \}$? How do I get this from the Fourier transforms table?

The given solution is $\displaystyle \sqrt{2} e^{\frac{4}{9}} e^{-i \frac{4}{3} x - x^2}$.

Complete the square of the exponent then use the appropriate translation theorem and the inverse transform of $\displaystyle e^{-w^2}$

CB
• Sep 30th 2011, 05:22 AM
math2011
Re: Inverse fourier transform using tables
Thank you. That worked.

$\displaystyle e^{-\frac{k^2}{4} + \frac{2}{3} k} = e^{-(x - \frac{4}{3})^2 + \frac{4}{9}}$
$\displaystyle \mathcal{F}^{-1} \{e^{-(x - \frac{4}{3})^2 + \frac{4}{9}} \} = \sqrt{2} e^{\frac{4}{9}} e^{-x^2} e^{-\frac{4}{3} i x}$